Is $H/L$ isomorphic to $K^*times K^*$ where $K^*=(K_{ne0},cdot)$ with $K$ field and $H, L$ certain subgroups...
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Let $K$ be a field, $H=left{begin{bmatrix}a&b\0&dend{bmatrix}:a,b,din K, adne0right}$, $L=left{begin{bmatrix}1&b\0&1end{bmatrix}:bin Kright}.$
I'm asked to prove that $H/L$ is isomorphic to $K^*times K^*$, where "$K^*=(K_{ne0},cdot)$". Now, this means I should find a bijective $f:H/Lto K^*times K^*$ such that $f(AB)=f(A)cdot f(B)$ - my first doubt is that the operator "$cdot$" is associated with the sole $K_{ne0}$ rather than $K_{ne0}times K_{ne0}$. Now, given that $H/L$ is isomorphic to $U=left{begin{bmatrix}a&0\0&dend{bmatrix}:a,din K, ane d, adne0right}$, am I correct in saying that $f:begin{bmatrix}a&0\0&dend{bmatrix}mapsto(a,d)$ does the job because $$fleft(begin{bmatrix}a_1&0\0&d_1end{bmatrix} begin{bmatrix}a_2&0\0&d_2end{bmatrix}right)=fleft(begin{bmatrix}a_1a_2&0\0&d_1d_2end{bmatrix}right)=(a_1a_2,d_1d_2)=(a_1,d_1)cdot(a_2,d_2)$$? I also have another concern: it seems to me that the $ane d$ bit in $U$ makes the cardinality of $U$ shift from that of $K_{ne0}times K_{ne0}$ (at least if $K$ is finite), so I guess it shouldn't be there? But why?
abstract-algebra matrices group-theory proof-verification group-isomorphism
asked Nov 17 at 14:20
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Let $K$ be a field, $H=left{begin{bmatrix}a&b\0&dend{bmatrix}:a,b,din K, adne0right}$, $L=left{begin{bmatrix}1&b\0&1end{bmatrix}:bin Kright}.$
I'm asked to prove that $H/L$ is isomorphic to $K^*times K^*$, where "$K^*=(K_{ne0},cdot)$". Now, this means I should find a bijective $f:H/Lto K^*times K^*$ such that $f(AB)=f(A)cdot f(B)$ - my first doubt is that the operator "$cdot$" is associated with the sole $K_{ne0}$ rather than $K_{ne0}times K_{ne0}$. Now, given that $H/L$ is isomorphic to $U=left{begin{bmatrix}a&0\0&dend{bmatrix}:a,din K, ane d, adne0right}$, am I correct in saying that $f:begin{bmatrix}a&0\0&dend{bmatrix}mapsto(a,d)$ does the job because $$fleft(begin{bmatrix}a_1&0\0&d_1end{bmatrix} begin{bmatrix}a_2&0\0&d_2end{bmatrix}right)=fleft(begin{bmatrix}a_1a_2&0\0&d_1d_2end{bmatrix}right)=(a_1a_2,d_1d_2)=(a_1,d_1)cdot(a_2,d_2)$$? I also have another concern: it seems to me that the $ane d$ bit in $U$ makes the cardinality of $U$ shift from that of $K_{ne0}times K_{ne0}$ (at least if $K$ is finite), so I guess it shouldn't be there? But why?
abstract-algebra matrices group-theory proof-verification group-isomorphism
asked Nov 17 at 14:20
Learner
1467
> my first doubt is that the operator "$cdot$" is associated with the sole K≠0 rather than $K_{neq 0}times K_{neq 0}$. $cdot$ here denotes a generic group operation (in particular, the one on $K^*times K^*$ induced by the one on $K^*$). Where did you get $a neq d$ from? It's not anywhere else in the question, and makes $U$ not a group (it's missing the identity).
– user3482749
Nov 17 at 15:12
@user3482749 Ok so I was thinking of $cdot$ correctly; as for $ane d$, how stupid of me, I should have seen that... I somehow thought $ane d$ stemmed from the observation that the multiples of elements of $L$ - but now that I think of it, this would make the identity matrix equal to $0$, which isn't possible. Could you please give some hint? I'm so confused
– Learner
Nov 17 at 15:42
add a comment |
up vote
1
down vote
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up vote
1
down vote
favorite
Let $K$ be a field, $H=left{begin{bmatrix}a&b\0&dend{bmatrix}:a,b,din K, adne0right}$, $L=left{begin{bmatrix}1&b\0&1end{bmatrix}:bin Kright}.$
I'm asked to prove that $H/L$ is isomorphic to $K^*times K^*$, where "$K^*=(K_{ne0},cdot)$". Now, this means I should find a bijective $f:H/Lto K^*times K^*$ such that $f(AB)=f(A)cdot f(B)$ - my first doubt is that the operator "$cdot$" is associated with the sole $K_{ne0}$ rather than $K_{ne0}times K_{ne0}$. Now, given that $H/L$ is isomorphic to $U=left{begin{bmatrix}a&0\0&dend{bmatrix}:a,din K, ane d, adne0right}$, am I correct in saying that $f:begin{bmatrix}a&0\0&dend{bmatrix}mapsto(a,d)$ does the job because $$fleft(begin{bmatrix}a_1&0\0&d_1end{bmatrix} begin{bmatrix}a_2&0\0&d_2end{bmatrix}right)=fleft(begin{bmatrix}a_1a_2&0\0&d_1d_2end{bmatrix}right)=(a_1a_2,d_1d_2)=(a_1,d_1)cdot(a_2,d_2)$$? I also have another concern: it seems to me that the $ane d$ bit in $U$ makes the cardinality of $U$ shift from that of $K_{ne0}times K_{ne0}$ (at least if $K$ is finite), so I guess it shouldn't be there? But why?
abstract-algebra matrices group-theory proof-verification group-isomorphism
asked Nov 17 at 14:20
Learner
1467
Let $K$ be a field, $H=left{begin{bmatrix}a&b\0&dend{bmatrix}:a,b,din K, adne0right}$, $L=left{begin{bmatrix}1&b\0&1end{bmatrix}:bin Kright}.$
I'm asked to prove that $H/L$ is isomorphic to $K^*times K^*$, where "$K^*=(K_{ne0},cdot)$". Now, this means I should find a bijective $f:H/Lto K^*times K^*$ such that $f(AB)=f(A)cdot f(B)$ - my first doubt is that the operator "$cdot$" is associated with the sole $K_{ne0}$ rather than $K_{ne0}times K_{ne0}$. Now, given that $H/L$ is isomorphic to $U=left{begin{bmatrix}a&0\0&dend{bmatrix}:a,din K, ane d, adne0right}$, am I correct in saying that $f:begin{bmatrix}a&0\0&dend{bmatrix}mapsto(a,d)$ does the job because $$fleft(begin{bmatrix}a_1&0\0&d_1end{bmatrix} begin{bmatrix}a_2&0\0&d_2end{bmatrix}right)=fleft(begin{bmatrix}a_1a_2&0\0&d_1d_2end{bmatrix}right)=(a_1a_2,d_1d_2)=(a_1,d_1)cdot(a_2,d_2)$$? I also have another concern: it seems to me that the $ane d$ bit in $U$ makes the cardinality of $U$ shift from that of $K_{ne0}times K_{ne0}$ (at least if $K$ is finite), so I guess it shouldn't be there? But why?
abstract-algebra matrices group-theory proof-verification group-isomorphism
abstract-algebra matrices group-theory proof-verification group-isomorphism
asked Nov 17 at 14:20
Learner
1467
asked Nov 17 at 14:20
Learner
1467
asked Nov 17 at 14:20
Learner
1467
asked Nov 17 at 14:20
Learner
1467
asked Nov 17 at 14:20
Learner
1467
1467
> my first doubt is that the operator "$cdot$" is associated with the sole K≠0 rather than $K_{neq 0}times K_{neq 0}$. $cdot$ here denotes a generic group operation (in particular, the one on $K^*times K^*$ induced by the one on $K^*$). Where did you get $a neq d$ from? It's not anywhere else in the question, and makes $U$ not a group (it's missing the identity).
– user3482749
Nov 17 at 15:12
@user3482749 Ok so I was thinking of $cdot$ correctly; as for $ane d$, how stupid of me, I should have seen that... I somehow thought $ane d$ stemmed from the observation that the multiples of elements of $L$ - but now that I think of it, this would make the identity matrix equal to $0$, which isn't possible. Could you please give some hint? I'm so confused
– Learner
Nov 17 at 15:42
add a comment |
> my first doubt is that the operator "$cdot$" is associated with the sole K≠0 rather than $K_{neq 0}times K_{neq 0}$. $cdot$ here denotes a generic group operation (in particular, the one on $K^*times K^*$ induced by the one on $K^*$). Where did you get $a neq d$ from? It's not anywhere else in the question, and makes $U$ not a group (it's missing the identity).
– user3482749
Nov 17 at 15:12
@user3482749 Ok so I was thinking of $cdot$ correctly; as for $ane d$, how stupid of me, I should have seen that... I somehow thought $ane d$ stemmed from the observation that the multiples of elements of $L$ - but now that I think of it, this would make the identity matrix equal to $0$, which isn't possible. Could you please give some hint? I'm so confused
– Learner
Nov 17 at 15:42
> my first doubt is that the operator "$cdot$" is associated with the sole K≠0 rather than $K_{neq 0}times K_{neq 0}$. $cdot$ here denotes a generic group operation (in particular, the one on $K^*times K^*$ induced by the one on $K^*$). Where did you get $a neq d$ from? It's not anywhere else in the question, and makes $U$ not a group (it's missing the identity).
– user3482749
Nov 17 at 15:12
> my first doubt is that the operator "$cdot$" is associated with the sole K≠0 rather than $K_{neq 0}times K_{neq 0}$. $cdot$ here denotes a generic group operation (in particular, the one on $K^*times K^*$ induced by the one on $K^*$). Where did you get $a neq d$ from? It's not anywhere else in the question, and makes $U$ not a group (it's missing the identity).
– user3482749
Nov 17 at 15:12
@user3482749 Ok so I was thinking of $cdot$ correctly; as for $ane d$, how stupid of me, I should have seen that... I somehow thought $ane d$ stemmed from the observation that the multiples of elements of $L$ - but now that I think of it, this would make the identity matrix equal to $0$, which isn't possible. Could you please give some hint? I'm so confused
– Learner
Nov 17 at 15:42
@user3482749 Ok so I was thinking of $cdot$ correctly; as for $ane d$, how stupid of me, I should have seen that... I somehow thought $ane d$ stemmed from the observation that the multiples of elements of $L$ - but now that I think of it, this would make the identity matrix equal to $0$, which isn't possible. Could you please give some hint? I'm so confused
– Learner
Nov 17 at 15:42
add a comment |
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> my first doubt is that the operator "$cdot$" is associated with the sole K≠0 rather than $K_{neq 0}times K_{neq 0}$. $cdot$ here denotes a generic group operation (in particular, the one on $K^*times K^*$ induced by the one on $K^*$). Where did you get $a neq d$ from? It's not anywhere else in the question, and makes $U$ not a group (it's missing the identity).
– user3482749
Nov 17 at 15:12
@user3482749 Ok so I was thinking of $cdot$ correctly; as for $ane d$, how stupid of me, I should have seen that... I somehow thought $ane d$ stemmed from the observation that the multiples of elements of $L$ - but now that I think of it, this would make the identity matrix equal to $0$, which isn't possible. Could you please give some hint? I'm so confused
– Learner
Nov 17 at 15:42